EE 422G Notes: Chapter 6
Instructor: Zhang
Chapter 6 Applications of the Laplace Transform
Part One: Analysis of Network (6-2, 6-3)
1. Review of Resistive Network
1) Elements
2) Superposition
Page 6-1
EE 422G Notes: Chapter 6
Instructor: Zhang
3) KVL and KCL
4) Equivalent Circuits
Page 6-2
EE 422G Notes: Chapter 6
Instructor: Zhang
5) Nodal Analysis and Mesh Analysis
Mesh analysis
VS1  R1I1  R2 ( I1  I 2 )  R3 ( I1  I 2 )

VS1  R1I1  R4 I 2  VS 2
Solve for I1 and I2.
2. Characteristics of Dynamic Network
Dynamic Elements  Ohm’s Law: ineffective
1) Inductor
2) Capacitor
Page 6-3
EE 422G Notes: Chapter 6
Instructor: Zhang
3) Example (Problem 5.9):
Why so simple? Algebraic operation!
Dynamic Relationships (not Ohm’s Law) Complicate the analysis
Using Laplace Transform
1  I ( s ) i ( 1) (0) 
Vs ( s )  L[ (t )]  L( sI ( s )  i (0 ))  

  RI ( s )
C s
s 

 ( Ls ) I ( s ) 
1
I ( s )  RI ( s )
Cs
Define ‘Generalized Resistors’ (Impedances)
Z1 (s)  Ls
Z 2 (s) 
1
Cs
 Vs ( s )  Z1 ( s ) I ( s )  Z 2 ( s ) I ( s )  RI ( s )
 I (s) 
Vs ( s )
Z1 ( s )  Z 2 ( s )  R
As simple as resistive network!
Solution proposed for dynamic network:
All the dynamic elements  Laplace Trans. Models.
Page 6-4
EE 422G Notes: Chapter 6
Instructor: Zhang


superposit ion

KVL and KCL
 As Resistive Network

Equivalent circuit

Nodal analysis and mesh analysis 
Generalize d Ohms Law

Key: Laplace transform models of (dynamic) elements.
3. Laplace transform models of circuit elements.
1) Capacitor
Important: We can handle these two ‘resistive network elements’!
2) Inductor
Page 6-5
EE 422G Notes: Chapter 6
Instructor: Zhang
3) Resistor V(s) = RI(s)
4)Sources
5) Mutual Inductance (Transformers)
(make sure both i1 and i2 either away
or toward the polarity marks to make
the mutual inductance M positive.)
Circuit (not transformer) form:
di1
di
di
di
M 1 M 1 M 2
dt
dt
dt
dt
di
d
 ( L1  M ) 1  M (i1  i2 )
dt
dt
di
di
di
di
v2 (t )  M 1  M 2  L2 2  M 2
dt
dt
dt
dt
di
d
 M (i1  i2 )  ( L2  M ) 2
dt
dt
v1 (t )  L1
Page 6-6
EE 422G Notes: Chapter 6
Instructor: Zhang
Benefits of transform

Let’s write the equations from this
circuit form:
The Same
 Laplace transform model: Obtain it by using inductance model
Just ‘sources’ and ‘generalized resistors’ (impedances)!
4. Circuit Analysis: Examples
Key: Remember very little, capable of doing a lot
How: follow your intuition, resistive network
‘Little’ to remember: models for inductor, capacitor and mutual inductance.
Example 6-4: Find Norton Equivalent circuit
Page 6-7
EE 422G Notes: Chapter 6
Instructor: Zhang
Assumption: vc (0  )  0
*Review of Resistive Network
1) short-circuit current through the load: I s
2) Equivalent Impedance or Resistance Rs or Z s :
A: Remove all sources
B: Replace Z L by an external source
C: Calculate the current generated by the external source ‘point a’
D: Voltage / Current  Rs  Z s
*Solution
1) Find I s (s)  I sc
1
1
3
s3
 I ( s )  1   3I ( s )  (1  ) I ( s ) 
I ( s)
s
s
s
s
1
2
I ( s) 
 I sc  2 I ( s )  
s3
s3
2) Find Z s
Zs 
Vtest ( s )
2 I (s)
Vtest(s)
3I ( s )
0
s
 3
 (1 ) I ( s)    I ( s)
s
I ( s)  1  
Page 6-8
EE 422G Notes: Chapter 6
Instructor: Zhang
(Will I(s) be zero? We don’t know yet!)
3)
condition: 1 ohm = 3/s
or I(s) = 0
=>I(s) = 0 =>Zs = 
a
ZL
Example 6-5: Loop Analysis (including initial condition)
Question: What are i0 and v0?
What is vc (t ) ?
Solution
1) Laplace Transformed Circuit
Why this direction?
Why this direction?
Page 6-9
EE 422G Notes: Chapter 6
Instructor: Zhang
2) KVL Equations

( R1  Ls ) I 1 ( s)  ( R2 )( I 1 ( s)  I 2 ( s))  Vs ( s)  Li0

v
1

( I 2 ( s)  I 3 ( s))   0
 R2 ( I 1 ( s)  I 2 ( s)) 
Cs
s

v0
 1

(
I
(
s
)

I
(
s
))

(
R

R
I
(
s
)

 VC ( s)
3
3
4) 3
 Cs 2
s
Important: Signs of the sources!
3) Simplified (Standard form)
Z ( s) I ( s)  E ( s)  V ( s)

 R1  R2  Ls

 R2



0

 R2
1
R2 
Cs
1

Cs

  I (s) 
 1 
  I 2 ( s)
  I (s) 
1
 3 
 R3  R4 

Cs
0
1

Cs




Vs ( s )  Li0  V ( s )   Li0 

  s   v 
v0
 
   0    0 
s

 
 s

V
(
s
)
 v0  V ( s)   C   v0 
C
 s

 s 
6-4 Transfer Functions
1. Definition of a Transfer Function
(1) Definition
System analysis : Emphasize relationsh ip between input and output, using blocks.

Network (Circuit) analysis : Details, examples : Branch currents, voltages.
Page 6-10
EE 422G Notes: Chapter 6
Instructor: Zhang
System analysis: How the system processes the input to form the output, or
Input : variable used and to be adjusted
to change or influence the output.
Can you give some examples for input and output?
Quantitative Description of ‘ how the system processes
the input to form the output’: Transfer Function H(s)
(2)  input
The resultant output y(t) to  (t) input: unit impulse response
In this case: X(s) = L [ (t)] = 1
Y(s) = Laplace Transform of the unit impulse response
=> H(s) = Y(s)/X(s) = Y(s)
Therefore: What is the transfer function of a system?
Answer : It is the Laplace transform of the unit impulse response
of the system.
(3) Facts on Transfer Functions
* Independent of input, a property of the system structure and parameters.
* Obtained with zero initial conditions.
(Can we obtain the complete response of a system based on its transfer
function and the input?)
* Rational Function of s (Linear, lumped, fixed)
* H(s): Transfer function
H( j2f ) or H( j ): frequency response function of the system
(Replace s in H(s) by j2f or j)
|H( j2f )| or |H( j )|: amplitude response function
Page 6-11
EE 422G Notes: Chapter 6
Instructor: Zhang
H(j2f) or H( j ): Phase response function
2. Properties of Transfer Function for Linear, Lumped stable systems
(1) Rational Function of s
Lumped, fixed, linear system =>
bm s m  bm 1s m 1  ...  b0 N ( s )
H ( s) 

n
n 1
an s  an 1s  ...  a0 D( s )
Corresponding differential equations:
d ( n ) y (t )
d ( n 1) y (t )
an
 a n 1
 ...  a0 y (t )
dt n
dt n 1
d ( m ) x(t )
d ( m 1) x(t )
 bm

b
 ...  b0 x(t )
m 1
dt m
dt m 1
(2)
aj
 all real! Why? Results from real system components.
bj 
Roots of N(s), D(s): real or complex conjugate pairs.
Poles of the transfer function: roots of D(s)
Zeros of the transfer function: roots of N(s)
Example:
D( s )  ( s 2  3s  1)  D( s )  ( s  2)( s  1)
poles : 2,1
(3) H(s) = N(s)/D(s) of bounded-input bounded-output (BIBO) stable
system
* Degree of N(s)  Degree of D(s)
Why? If degree N(s) > Degree D(s)
 H ( s )  ck s k  ...  c1s  c0 
N ' ( s)
D( s)
where degree N’(s) <degree D(s)
Under a bounded-input x(t) = u(t) => X(s) = 1/s
Page 6-12
EE 422G Notes: Chapter 6
Instructor: Zhang
 Y ( s )  ck s k 1  ...  c1 
c0 N ' ( s )

s sD( s )
(  (t ) not bounded!)
 y (t )  ...  c1 (t )  ...
* Poles: must lie in the left half of the s-plan (l. h. p)
i.e.,
D( s)  ( s   1 )( s   2 )...( s   n )
Re( j )  0
Why?
Y ( s )  ... 
Bj
(s   j ) k
 y (t )  ... 
 ... 
B j t k 1
(k  1)!
B j t k 1
(k  1)!
 ...( j   j  j )
e
 jt
 ...
 t
e j e jt  ...
(Can we also include k=1 into this form? Yes!)
B j t k 1
(k  1)!
 t
e j e j t
 t
k  1  B j e j e jt
Magnitude : | e jt | 1, | B j |: fixed
 t
B j e j e jt | B j || e jt | e
| B j | e
 jt
 jt
0
j 0
 cons tan t

j 0
j 0
* Any restriction on zeros? No (for BIBO stable system)
3. Components of System Response
d ( n ) y (t )
d ( m ) x (t )
an
 ...  a0 y (t )  bm
 ...  b0 x(t )
dt n
dt m
Because x(t) is input, we can assume
x(0)  x(1) (0)  ...  x( m1) (0)  0
Page 6-13
EE 422G Notes: Chapter 6
Instructor: Zhang
Laplace transform of the differentional equation
an s nY ( s )  an [ s n1 y (0)  s n2 y (1) (0)  ...  y ( n1) (0)
 an1s n1Y ( s )  an1[ s n2 y (0)  ...  y ( n2 ) (0)
D(s)
 ...

a1sY ( s )  a1 y (0)

a0Y ( s )
C(s)
 [bm s m  bm1s m1  ...  b0 ] X ( s )
 D( s )Y ( s )  C ( s )  N ( s ) X ( s )
or Y ( s ) 
N(s)
C (s) N (s)
C ( s)

X ( s) 
 H ( s) X ( s)
D( s) D( s)
D( s)
D(s): System parameters
C(s): Determined by the initial conditions (initial states)
Initial-State Response (ISR) or Zero-Input Response (ZIR):
 C ( s) 
y zir (t )  L1 

 D( s ) 
Zero-State Response (ZSR) (due to input)
yzsr (t )  L1[ H ( s) X ( s)]
From another point of view:
Transient Response: Approaches zero as t∞
Forced Response: Steady-State response if the forced
response is a constant
How to find (1) zero-input response or initial-state response? No problem!
 C ( s) 
L1 

 D( s ) 
(2) zero-state response? No prolbem!
L1[ H ( s) X ( s)]
Page 6-14
EE 422G Notes: Chapter 6
Instructor: Zhang
How to find (1) transient response? All terms which go to 0 as t
(2) forced response? All terms other than transient terms.
Example 6-7
Input v s (t )
Output y(t )  vR (t )
Initial capacitor voltage: v0
RC = 1 second
Solution
(1) Find total response
Vs ( s )  v0 / s
R
1 / Cs  R
Rv 0 / s
R


Vs ( s )
1 / Cs  R 1 / Cs  R
v0 / s
1


Vs ( s )
1 / CRs  1 1 / CRs  1
v0
s


Vs ( s )
1 / CR  s 1 / CR  s
CR 1
v
s
  0 
Vs ( s )
s 1 s 1
Y ( s) 
(2) Find zero-input response and zero-state response
v0 
1 
 v0 e t u (t )
Zero-input response: L 

 s  1
Zero-state response:
 s  5s  
 s

L1 
Vs ( s )  L1 
 2

s 1

 s  1  s  4 
 1
1 2 
 s
 L1 
 4 2


2
 s  1  s  4 2 s  4 
 [e t  4 cos 2t  2 sin 2t ]u (t )
(3) Find transient and forced response
y (t )  v0 e t u (t )  e t u (t )  4(cos 2t )u (t )  2(sin 2t )u (t )
Which terms go to zero as t?
Page 6-15
EE 422G Notes: Chapter 6
Instructor: Zhang
 v0 e t u(t )
and
e t u (t )
 ytransient (t )  v0 e t u (t )  e t u(t )
What are the other terms:
4(cos 2t )u (t )
and
 2(sin 2t )u (t )
 y forced (t )  (4 cos 2t  2 sin 2t )u (t )
4. Asymptotic and Marginal Stability
System: (1) Asymptotically stable if y zir  0 as t (no input) for all
possible initial conditions, y(0), y’(0), … y(n-1)(0)
 Internal stability, has nothing to do with external input/output
(2) Marginally stable
| y zir  M
all t>0 and all initial conditions
(3) Unstable
y zir grows without bound for at least some values
of the initial condition.
(4) Asymptotically stable (internally stable)
=>must be BIBO stable. (external stability)
6-5 Routh Array
1. Introduction
System H(s) = N(s)/D(s) asymptotically stable  all poles in l.h.p (not
include jw axis.
How to determine the stability?
Factorize D(s):
D( s )  s n  an 1s n 1  ...  a1s  a0
 ( s  1 )( s   2 )...( s   n )
stable  all
Re( j )  0
Page 6-16
EE 422G Notes: Chapter 6
Instructor: Zhang
Other method to determine (just) stability without factorization?
Routh Array
(1)
Necessary condition
All a j ( j  0,1,...n  1)  0
(when an  1 is used)
or
a j  0 => system unstable!
 any a j  0
Why?
Denote  j   p j  Re( p j )  0 to esnure stability
D( s )  ( s  p1 )( s  p 2 )...( s  p n )
 sum of products of p j ' s 

 s n  ( p1  p 2  ... p n ) s n 1  s n  2 

taken
two
at
a
time


 sum of products of p j ' s 
 s  p1 p 2 ... p n
 ...  

taken
n1
at
a
time


When all Re(pj) > 0 , all coefficients must be greater than zero. If some
coefficient is not greater than zero, there must
be at least a Re(pj) <= 0 (i.e., Re(  j )  Re( Pole j )  0 )
=> system unstable
(2)
Routh Array
Question: All a j  0 implies system stable?
Not necessary
Judge the stability: Use Routh Array (necessary and sufficient)
2. Routh Array Criterion
Find how many poles in the right half of the s-plane
(1) Basic Method
D( s)  an s n  an 1s n 1  ...  a0
Page 6-17
EE 422G Notes: Chapter 6
Instructor: Zhang
Formation of Routh Array
Number of sign changes in the first column of the array
=> number of poles in the r. h. p.
Example 6-8
D( s)  s3  14s 2  41s  56
sign: Changed once =>one pole in the r.h.p
verification:
D( s )  s 3  14 s 2  41s  56
 ( s  1)( s  7)( s  8)
Example 6-9
D( s)  s 4  5s3  s 2  10s  1
Page 6-18
EE 422G Notes: Chapter 6
Instructor: Zhang
5  1  1  10
 1
5
5 1  1 0
b2 
1
5
 1  10  5  1
c1 
 15
1
 1 0  5  0
c2 
0
1
15  1  ( 1)  0
d1 
1
15
b1 
s4
s3
s2
s1
s0
1
5
1
10
1
b1  1 b2  1
c1  15 c2  0
d1  1
Sign: changed twice => two poles in r.h.p.
(2) Modifications for zero entries in the array
Case 1: First element of a row is zero
 replace 0 by ε (a small positive number)
Example 6-10
D( s)  s 4  s3  s 2  s  3
Case 2: whole row is zero (must occur at odd power row)
construct an auxiliary polynomial and the perform differentiation
Example: best way.
Page 6-19
EE 422G Notes: Chapter 6
Instructor: Zhang
Example 6-11
D( s)  s7  3s6  3s5  s 4  s3  3s 2  3s  1
S
S3
Replace 0
Page 6-20
EE 422G Notes: Chapter 6
Instructor: Zhang
(3) Application: Can not be replaced by MatLab
Range of some system parameters.
Example:
D( s)  s3  3s 2  3s  (1  k )
s3
1
3
s2
3
1 k
s1
b1  (8  k ) / 3
s0
c1  1  k
3  3  1  (1  k ) 8  k

3
3
(8  k ) / 3  (1  k )  3  0
c1 
 1 k
(8  K ) / 3
b1 
Stable system
8k

08 k 
3
  8  k  1
1  k  0  k  1

to ensure system stable!
6-6 Frequency Response and Bode Plot
Transfer Function H ( s ) 
Frequency Response
N ( j )
H ( j ) 
D( j )
N ( s)
D( s)
H ( j ) 
Amplitude Response: | H ( j ) |
H ( s) 
2
s 2  3s  1
2
2

( j ) 2  3 j  1 (1   2 )  j (3 )
2
Why?
(1   2 )2  32  2
Real positive number: function of
Phase Response: H ( j )   tan 1

3
12
Page 6-21
EE 422G Notes: Chapter 6
Instructor: Zhang
Interest of this section
In particular, obtain
What are these?
Page 6-22
EE 422G Notes: Chapter 6
Instructor: Zhang
Important Question: What is a Bode Plot?
How to obtain them without much computations?
Asymptotes only!
1. Bode plots of factors
(1) Constant factor k:
(2) s
s 1 :
| H |dB  20 log10 |
1
| 20 log10 | j | 20 log10 
j
Page 6-23
EE 422G Notes: Chapter 6
Instructor: Zhang
Can we plot it?
Can we plot | H |dB ~  for them?
S N :
Phase s: H ( j )  90o
S N :
(3)
H ( j )   N  90o
1
Ts  1
step 1: Coordinate systems
step 2: corner frequency
c 
1
T
step 3: Label 0.1c, c , 10c
step 4: left of c : | H |dB  0
step 5: right of c :
(   c ,
(  10 c ,
Point 1
| H |dB  0)
| H |dB  20dB )
Line
Point 2
Page 6-24
EE 422G Notes: Chapter 6
Why?
Instructor: Zhang
| H ( j ) |dB  20 log10 | H ( j ) |
 20 log10 |
1
1
|s  j  20 log10 |
|
Ts  1
1  jT
 20 log10 | 1  jT | 20 log10 1   2T 2
If
T  1
( 
1
 c )
T
 1   2T 2  1 | H ( j ) |dB  20 log10 1  0
If
T  1
( 
1
 c )
T
 1   2T 2  T | H ( j ) |dB  20 log10 T
10
 T  10  20 log10 10  20dB
T
T  100 c | H ( j ) |dB  20 log10 100  40dB
  10 c 
Example:
1
0 .2 s  1
What is T : T = 0.2
What is c : c = 1/T = 5
Example : 0.2s + 1
Example : (0.2s + 1)2,
(0.2s + 1)-2
Page 6-25
EE 422G Notes: Chapter 6
Instructor: Zhang
Example : (Ts + 1)±N
(4) (T 2 s 2  2Ts  1)1 (Complex --- Conjugate poles)
Step 3 : Before   c : | H |dB  0
Right of   c :
point 1: (   c , | H |dB  0 )
Line
point 2: (   10c , | H |dB  40dB )
Example: (0.2 2 s 2  2  0.2s  1) 1
Actual | H |dB ~  and  (show Fig 6-20)
Page 6-26
EE 422G Notes: Chapter 6
Instructor: Zhang
What’s resonant frequency: reach maximum: c 1  2 2
Under what condition we have a resonant frequency:
  2 / 2  0.707
H ( j ) ~  : see fig 6-21
What about : (T 2 s 2  2Ts  1) N ?
2. Bode plots: More than one factors
Can we sum two | H |dB ~  plots into one?
Can we sum two H ( j ) ~  plots into one?
Yes!
Page 6-27
EE 422G Notes: Chapter 6
Instructor: Zhang
3. MatLab
H ( s) 
100( s  2)
100 s  200

s( s 2  5s  25) s 3  5s 2  25s
 Num  0 0 100 200;
 den  1 5 25 0;
 bode( num, den )
Show result in fig 6-24
6.7 Block Diagrams
1. What is a block diagram?
Concepts: Block, block transfer function,
Interconnection, signal flow, direction
Summer
System input, system output
Simplification, system transfer function
2. Block
Page 6-28
EE 422G Notes: Chapter 6
Instructor: Zhang
Assumption: Y(s) is determined
by input (X(s)) and block transfer
function (G(s)). Not affected by
the load.
Should be vary careful in
analysis of practical systems about the accuracy of this assumption.
3. Cascade connection
4. Summer
5. Single-loop system
Page 6-29
EE 422G Notes: Chapter 6
C ( s)
?
R( s )
Instructor: Zhang
C ( s)
 G( s)
R( s )
or
C ( s )  G ( s ) R( s ) ?
N 0!
C ( s )  G ( s ) E ( s )!
T ( s )  C ( s ) / R( s )
Let’s find
Closed-loop transfer function
C ( s)  G( s) E ( s)
Equation (1)
E ( s )  R( s )  H ( s )C ( s )
Equation (2)
C ( s )  G ( s )( R ( s )  H ( s )C ( s ))
 G ( s ) R ( s )  G ( s ) H ( s )C ( s )
 (1  G ( s ) H ( s ))C ( s )  G ( s ) R ( s )
 T ( s)  C ( s) / R( s) 
G( s)
1  G( s) H ( s)
6. More Rules and Summary: Table 6-1
Page 6-30
EE 422G Notes: Chapter 6
Instructor: Zhang
Page 6-31
EE 422G Notes: Chapter 6
Instructor: Zhang
Example 6-14: Find Y(s)/X(s)
G1 (G3  G2G4 )
Y ( s)
G1 (G3  G2G4 )
1  G1H1


X ( s ) 1  G1 (G3  G2G4 ) H
1  G1H1  H 2G1 (G3  G2G4 )
2
1  G1H1
Example 6-15: Armature- Controlled dc servomotor
Input : Ea (armature voltage)
Output :  (angular shift)
Page 6-32
EE 422G Notes: Chapter 6
Instructor: Zhang
Can we obtain  / E a ?
Example 6-16 Design of control system
Design of K such that closed loop system stable.
k ( s  1 / 2) 5
o ( s)
5k ( s  0.5)
5k ( s  0.5)
s
s( s  1)

 2
 3
 i ( s ) 1  k ( s  1 / 2) 5
s ( s  1)  5k ( s  0.5) s  s 2  5ks  2.5k
s
s( s  1)
Routh Array:
s3  s 2  5ks  2.5k
s3
1
5k
s2
1
2.5k
1
b1  2.5k
b2  0
0
c1  2.5k
s
s
5k  2.5k
 2.5k
1
b2  0
b1 
c1 
2.5k 2.5k  0
 2.5k
2.5k
System stable if k>0. If certain performance is required in addition to the
stability, k must be further designed.
Page 6-33